f(x)=sin(2x-π/6),x∈[π/4,5π/6] 求值域
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f(x)=sin(2x-π/6),x∈[π/4,5π/6] 求值域
f(x)=sin(2x-π/6),x∈[π/4,5π/6] 求值域
f(x)=sin(2x-π/6),x∈[π/4,5π/6] 求值域
x∈[π/4,5π/6] ,(2x-π/6)∈[π/3,3π/2] ,f(x)=sin(2x-π/6)∈[-1,1]
因为,x∈[π/4,5π/6]
所以 2x-π/6 属于[π/3,3π/2]
视2x-π/6为t
则求sint的值域
为[-1,1]
X∈[π/4,5π/6],所以2X-π/6∈[π/3,3π/2],所以按照正弦曲线看sin(2x-π/6)∈[-1,1]
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